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C H A P T E R 1Limits and Their PropertiesSection 1.1A Preview of Calculus . . . . . . . . . . . . . . . . . . . . 46Section 1.2Finding Limits Graphically and Numerically . . . . . . . . 47Section 1.3Evaluating Limits AnalyticallySection 1.4Continuity and One-Sided Limits . . . . . . . . . . . . . . 68Section 1.5Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . 78. . . . . . . . . . . . . . . 57Review Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85Problem Solving. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
C H A P T E R 1Limits and Their PropertiesSection 1.1A Preview of Calculus1. Precalculus: 20 ft sec 15 seconds 300 feet2. Calculus: velocity is not constant3. Calculus required: slope of tangent line at x 2 is rate ofchange, and equals about 0.16.4. Precalculus: rate of change slope 0.0811155. Precalculus: Area 2 bh 2 5 3 2 sq. units6. Calculus required: Area bhDistance 20 ft sec 15 seconds 300 feet 2 2.5 5 sq. units7. f x 4x x2(a)8. f x xy(a)yP(4, 2)432Px1x1223453(b) slope m 4x x2 3x 1(b) slope m x 1 3 x 3 x, x 1x 1 x 2: m 3 2 1x 1: m x 1.5: m 3 1.5 1.5x 0.5: m 3 0.5 2.5x 3: m (c) At P 1, 3 the slope is 2.x 5: m You can improve your approximation of the slope atx 1 by considering x-values very close to 1. x 2x 4 x 2 x 2 x 21 1 21 3 21 5 2(c) At P 4, 2 the slope is 1 x 2, x 413 0.2679 0.23611 4 2 1 0.25.4You can improve your approximation of the slope atx 4 by considering x-values very close to 4.5 5 5 10.4172 3 415555555Area 5 9.14521.5 2 2.5 3 3.5 4 4.59. (a) Area 5 (b) You could improve the approximation by using more rectangles.46
Section 1.210. (a) For the figure on the left, each rectangle has widthArea 3 sin sin sin sin4424 2 2 1 4 22 2 14 .4 1.8961For the figure on the right, each rectangle has widthArea .6 2 5 sin sin sin sin sin663236 3 1 31 1 6 2222 3 26Finding Limits Graphically and Numerically 1.9541(b) You could obtain a more accurate approximation by using more rectangles. You will learn later that the exact area is 2.11. (a) D1 5 1 2 1 5 2 16 16 5.66555555(b) D2 1 2 1 2 3 1 3 4 1 4 1 2222 2.693 1.302 1.083 1.031 6.11(c) Increase the number of line segments.Section 1.21.x1.91.991.9992.0012.012.1f x 0.34480.33440.33340.33320.33220.3226limx 22.x 2 0.3333x2 x 2 Actual limit is 13 . x1.91.991.9992.0012.012.1f x 0.25640.25060.25010.24990.24940.2439limx 23.Finding Limits Graphically and Numericallyx 2 0.25x2 4 0.1xf x limx 0 Actual limit is 14 . 0.010.29110.2889 x 3 3x 0.001 0.28870.28870.0010.010.10.28870.28840.2863 Actual limit is 1 2 3 . 47
484.Chapter 1x 3.1 3.01 3.001 2.999 2.99 2.9f x 0.2485 0.2498 0.2500 0.2500 0.2502 0.2516lim 1 x 2xlimx 32.9993.0013.01 0.0641 0.0627 0.0625 0.0625 0.0623 1 x 1 1 4 0.0625x 33.993.9994.0014.014.1f x 0.04080.04010.04000.04000.03990.0392 x x 1 4 5 0.04x 4 0.1xf x limx 00.9983f x limx 00.99998sin x 1.0000x 0.1x 0.010.0500 0.0610 Actual limit is 251 . 0.0011.00000.0010.010.11.00000.999980.9983(Actual limit is 1.) (Make sure you use radian mode.) 0.010.0050cos x 1 0.0000x 0.0010.00050.0010.01 0.0005 0.00500.1 0.0500(Actual limit is 0.) (Make sure you use radian mode.)9. lim 4 x 1x 311. lim f x lim 4 x 2x 23.1 Actual limit is 161 . 3.9lim8.2.99xx 47. Actual limit is 14 . 2.9f x 6. 0.25x 3x 35.Limits and Their Propertiesx 2x 5does not exist. For values of x to the left of 5,x 5 x 5x 5 x 5 equals 1, whereas for values of x to theright of 5, x 5 x 5 equals 1.10. lim x2 2 3x 112. lim f x lim x2 2 3x 1x 11does not exist since the function increases andx 3decreases without bound as x approaches 3.13. lim14. lim15. lim sin x 016. lim sec x 117. lim cos 1 x does not exist since the function oscillates18. lim tan x does not exist since the function increases andx 1x 0between 1 and 1 as x approaches 0.x 3x 0x 2decreases without bound as x approaches 2.
Section 1.219. (a) f 1 exists. The black dot at 1, 2 indicates thatf 1 2.Finding Limits Graphically and Numerically49(c) f 4 does not exist. The hollow circle at 4, 2 indicates that f is not defined at 4.(b) lim f x does not exist. As x approaches 1 from the(d) lim f x exists. As x approaches 4, f x approaches 2:x 1x 4left, f x approaches 2.5, whereas as x approaches 1from the right, f x approaches 1.lim f x 2.x 4(e) f 2 does not exist. The hollow circle at 2, 12 indicates that f 2 is not defined.20. (a) f 2 does not exist. The vertical dotted line indicatesthat f is not defined at 2.(b) lim f x does not exist. As x approaches 2, the1(f) lim f x exists. As x approaches 2, f x approaches 2 :x 2x 2values of f x do not approach a specific number.lim f x 12.x 2(c) f 0 exists. The black dot at 0, 4 indicates thatf 0 4.(g) f 4 exists. The black dot at 4, 2 indicates thatf 4 2.(d) lim f x does not exist. As x approaches 0 from the(h) lim f x does not exist. As x approaches 4, the valuesx 0left, f x approaches 12, whereas as x approaches 0from the right, f x approaches 4.x 4of f x do not approach a specific number.21. lim f x exists for all c 3. In particular, lim f x 2.22. lim f x exists for all c 2, 0. In particular, lim f x 2.23.24.x cx 2yx cx 4y625413f211234π2 π2x 2 1 15xπ 1 2lim f x exists for all values of c 4.lim f x exists for all values of c .x cx cy25. One possible answer isy26. One possible answer is64534f2211 2 1 127. C t 0.75 0.50 t 1 (a) 3x1234 2x 112 15(b)t33.33.43.53.63.74C 1.752.252.252.252.252.252.253lim C t 2.25t 2.25lim C t does not exist. The values of C jump from 1.75 to 2.25 at t 3.t 3
50Chapter 1Limits and Their Properties28. C t 0.35 0.12 t 1 (a)1050(b)tC t 33.33.43.53.63.740.590.710.710.710.710.710.71lim C t 0.71t 3.5(c)tC t 22.52.933.13.540.470.590.590.590.710.710.71lim C t does not exist. The values of C jump from 0.59 to 0.71 at t 3.t 329. We need f x 3 x 1 3 x 2 0.4. Hence, take 0.4. If 0 x 2 0.4, thenx 2 x 1 3 f x 3 0.4, as desired.12 x11 1 0.01. Let . If 0 x 2 , thenx 1x 110110130. We need f x 1 1111 x 2 1 x 1 1 101101101101 100102 x 1 101101 x 1 100101and we have112 x1 101 1 0.01. x 1x 1100 101 100f x 1 31. You need to find such that 0 x 1 implies1f x 1 1 0.1. That is,x 0.1 1 1 0.1x1 0.1 1x 1 0.19 101x 1110 101110 9x1010 1 x 1 191111 x 1 .911So take 1. Then 0 x 1 implies11 11 x 1 1111 11 x 1 .119Using the first series of equivalent inequalities, you obtainf x 1 1 1 0.1.x
Section 1.232. You need to find such that 0 x 2 impliesf x 3 x2 1 3 x2 4 0.2. That is, 0.24 0.23.8 3.8 3.8 2x2 x 4x2x2x 2 0.24 0.24.2 4.2 4.2 2So take 4.2 2 0.0494.Then 0 x 2 impliesFinding Limits Graphically and Numerically33. lim 3x 2 8 Lx 2 3x 2 8 0.013x 6 0.013 x 2 0.010.01 0.0033 30.01Hence, if 0 x 2 , you have30 x 2 3 x 2 0.01 4.2 2 x 2 4.2 23x 6 0.01 3.8 2 x 2 4.2 2. 3x 2 8 0.01Using the first series of equivalent inequalities, you obtainf x L 0.01f x 3 x2 4 0.2. x 2 L234. lim 4 x 44 x 2 0.0122 x 0.0121 x 4 0.0120 x 4 0.02 Hence, if 0 x 4 0.02, you have1 x 4 0.0122 4 x 0.012 x 2 0.012f x L 0.0136. lim x2 4 29 Lx 5 x2 4 29 0.01x 25 0.012 x 5 x 5 0.01x 5 0.01x 5If we assume 4 x 6, then 0.01 11 0.0009.5135. lim x2 3 1 Lx 2 x2 3 1 0.01x2 4 0.01 x 2 x 2 0.01x 2 x 2 0.01x 2 0.01x 2If we assume 1 x 3, then 0.01 5 0.002.Hence, if 0 x 2 0.002, you have11x 2 0.002 0.01 0.01 5x 2x 2 x 2 0.01x2 4 0.01 x2 3 1 0.01f x L 0.01Hence, if 0 x 5 x 5 0.01, you have110.011 0.01 11x 5x 5 x 5 0.01x2 25 0.01 x2 4 29 0.01f x L 0.01
52Chapter 1Limits and Their Properties37. lim x 3 538. lim 2x 5 1x 2x 3Given 0:Given 0: x 3 5 x 2 2x 5 1 2x 6 Hence, let .2x 3 Hence, if 0 x 2 , you havex 3 x 2 2Hence, let 2. x 3 5 Hence, if 0 x 3 , you have2f x L .x 3 22x 6 2x 5 1 f x L .39. limx 4 12 x 1 12 4 1 3Given 0:x 1Given 0: 12x 1 3 12x1240. lim 23 x 9 23 1 9 293 23x 9 293 23x 2 23x 4 23 x 1 x 1 32 x 4 2 Hence, let 2 .Hence, let 3 2 .Hence, if 0 x 4 2 , you haveHence, if 0 x 1 32 , you havex 1 32 x 4 2 12x23x 2 12x 1 3 23 23x 9 293 f x L .f x L .42. lim 1 141. lim 3 3x 6x 2Given 0:Given 0: 1 1 3 3 0 Hence, any 0 will work.Hence, for any 0, you have3 3 f x L .0 Hence, any 0 will work.Hence, for any 0, you have 1 1 f x L .
Section 1.2343. lim x 0Finding Limits Graphically and Numerically44. lim x 4 2x 0Given 0: 3 x x x 4Given 0: 0 3 x 3 x 2 x 2 x 2 x 2 . 45. lim x 2 2 2 4 46. lim x 3 0x 3Given 0:Given 0: x 2 4 x 2 4 x 2 x 2 x 2 x 2 0 x 3 0 x 3 Hence, let . Hence for 0 x 3 , you haveHence, let . Hence for 0 x 2 , you have x 2 x 2 x 2 4 x 2 4 f x L . x 3 x 3 0 f x L .(because x 2 0)47. lim x2 1 248. lim x2 3x 0x 1x 3Given 0:Given 0: x2 3x 0 x x 3 x2 1 2 x2 1 x 1 x 1 x 1 x 1 If we assume 0 x 2, then 3. Hence for 0 x 1 , you have3 11 x 1 3 x 1 x2 1 x2 1 2 f x 2 . 0 x 4 3 x 4 x 2 x 3 3 x 3 x 0 f x L .Hence for 0 x 0 3, you have x 4 x 2 Assuming 1 x 9, you can choose 3 . Then,Hence, let 3.x 2 x 2 53 x 3 x If we assume 4 x 2, then 4. Hence for 0 x 3 , you have4 11 x 3 4 x x x 3 x2 3x 0 f x L .
54Chapter 149. f x Limits and Their Properties x 5 30.550. f x x 4lim f x x 416 6x2x 3 4x 3lim f x 6x 34 312 0.1667The domain is 5, 4 4,not show the hole at 4, 16 .51. f x x 9 4 . The graphing utility doesThe domain is all x 1, 3. The graphing utility does notshow the hole at 3, 12 .52. f x 10 x 3lim f x 6x 3x2 9lim f x x 9x 30316 9310 30The domain is all x 3. The graphing utility does not1show the hole at 3, 6 .The domain is all x 0 except x 9. The graphingutility does not show the hole at 9, 6 .53. lim f x 25 means that thex 8values of f approach 25 as x getscloser and closer to 8.54. No. The fact that f 2 4 has nobearing on the existence of thelimit of f x as x approaches 2.56. (i) The values of f approach differentnumbers as x approaches c fromdifferent sides of c:y55. No. The fact that lim f x 4 hasx 2no bearing on the value of f at 2.(ii) The values of f increase without bound as x approaches c:(iii) The values of f oscillatebetween two fixed numbers asx approaches c:yy64534234321 4 3 2 1 1x12341 3 2 1 13425 3 4 4458. V r 3, V 2.48357. (a) C 2 rC63 0.9549 cm2 2 (b) If C 5.5, r 5.5 0.87535 cm2 If C 6.5, r 6.5 1.03451 cm2 Thus 0.87535 r 1.03451(c) lim 2 r 6; 0.5; 0.0796r 3 x 4 3 2x2 2 3r 54(a) 2.48 r 33r3 1.86 r 0.8397 in.(b)2.45 2.45 V 2.514 3 r 2.5130.5849 r 3 0.59920.8363 r 0.8431(c) For 2.51 2.48 0.03, 0.00334
Section 1.259. f x 1 x 1 xlim 1 x 1 x e 2.71828x 0y73(0, 2.7183)21xf x xf x 0.12.8679720.12.593742 0.012.7319990.012.704814 0.0012.7196420.0012.716942 0.00012.7184180.00012.718146 0.000012.7182950.000012.718268 0.0000012.7182830.0000012.71828055x 3 2 1 160. f x Finding Limits Graphically and Numerically12345x 1 x 1xy3xf x 1 0.5 0.100.10.51.0222Undef.2221 2x 1lim f x 212 1x 0Note that for 1 x 1, x 0, f x 61. x 1 x 1 2.x62.0.0020.005(1.999, 0.001)(2.001, 0.001)1.99802.992.0023.010Using the zoom and trace feature, 0.001. That is, for0 x 2 0.001,x2 4 4 0.001.x 263. False; f x sin x x is undefined when x 0.From Exercise 7, we havelimx 0From the graph, 0.001. Thus 3 , 3 2.999, 3.001 .Note:x2 3x x for x 3.x 364. Truesin x 1.x65. False; letf x 66. False; let x 4x,10,2x 4.x 4f x f 4 10 10,x2 4x,x 4x 4.lim f x lim x2 4x 0 and f 4 10 0x 4x 4lim f x lim x2 4x 0 10x 4x 467. f x x(b) lim x 0 is false.(a) lim x 0.5 is true.x 0x 0.25As x approaches 0.25 12 0.5.14,f x x approachesf x x is not defined on an open interval containing 0because the domain of f is x 0.
56Chapter 1Limits and Their Properties68. The value of f at c has no bearing on the limit as x approaches c.69. If lim f x L1 and lim f x L2, then for every 0, there exists 1 0 and 2 0 such thatx cx cx c 1 f x L1 and x c 2 f x L2 . Let equal the smaller of 1 and 2.Then for x c , we have L1 L2 L1 f x f x L2 L1 f x f x L2 .Therefore, L1 L2 2 . Since 0 is arbitrary, it follows that L1 L2.70. f x mx b, m 0. Let 0 be given. Take .m , thenmIf 0 x c 71. lim f x L 0 means that for every 0 therex cexists 0 such that if0 x c ,m x c thenmx mc f x L 0 . mx b mc b This means the same as f x L whenwhich shows that lim mx b mc b.0 x c .x cThus, lim f x L.x c72. (a) 3x 1 3x 1 x2 0.01 9x2 1 x2 9x4 x2 11001(b) We are given lim g x L 0. Let 2L. Therex cexists 0 such that 0 x c implies thatLg x L . That is,211001 10x2 1 90x2 1 100 Thus, 3x 1 3x 1 x2 0.01 0 if10x2 1 0 and 90x2 1 0.Let a, b 1,1 90 90LL g x L 22L 2 .g x For x in the interval c , c , x c, we haveLg x 0, as desired.2For all x 0 in a, b , the graph is positive. You canverify this with a graphing utility.73. Answers will vary.74. limx 4x2 x 12 7x 4n4 0.1nf 4 0.1 n 3L2n4 0.1nf 4 0.1 n .99944.00017.000143.99996.9999
Section 1.375. The radius OP has a length equal to the altitude z of thehhtriangle plus . Thus, z 1 .22 1hArea triangle b 1 22 Evaluating Limits AnalyticallyPhOArea rectangle bhb 1hSince these are equal, b 1 bh221 h 2h25h 12h 2576. Consider a cross section of the cone, where EF is a diagonal of the inscribed cube.AD 3, BC 2. Let x be the length of a side of the cube. Then EF x 2.ABy similar triangles,EEFAG BC ADx 2 3 x 23BGDFC3 2x 6 2xSolving for x, 3 2 2 x 6x Section 1.31.63 2 2 9 2 6 0.96.7Evaluating Limits Analytically(a) lim h x 07x 5 8132.(a) lim g x 2.410x 4(b) lim g x 4(b) lim h x 6x 0x 10 7 5h x x2 5x3.10g x (a) lim f x 04x 0π π 4f x x cos x(b) lim f x 0.524x 3 6 12 x 3 x 94.(a) lim f t 010t 4(b) lim f t 5 510 10 f t t t 4t 157
58Chapter 1Limits and Their Properties5. lim x4 24 166. lim x3 2 3 87. lim 2x 1 2 0 1 18. lim 3x 2 3 3 2 7x 2x 2x 3x 09. lim x2 3x 3 2 3 3 9 9 010. lim x2 1 1 2 1 0x 3x 111. lim 2x2 4x 1 2 3 2 4 3 1 18 12 1 7x 312. lim 3x3 2x2 4 3 1 3 2 1 2 4 514. limx 316. limx 318. limx 322 2x 2 3 2x 4 3 13 41 1 x215. limx 31 3 22 x2 4 12 455x 12x 3 2 3 3 3 x 53 58 x 113. limx 2x 117. limx 75x x 2 5 7 7 2 35 9 35319. lim x 1 3 1 2 2x 33320. lim x 4 4 4 221. lim x 3 2 4 3 2 122. lim 2x 1 3 2 0 1 3 123. (a) lim f x 5 1 4x 4x 4x 0x 1(b) lim g x 43 64x 4(c) lim g f x g f 1 g 4 64x 124. (a) lim f x 3 7 425. (a) lim f x 4 1 3x 3x 1(b) lim g x 42 16(b) lim g x 3 1 2(c) lim g f x g 4 16(c) lim g f x g 3 2x 4x 3x 3x 126. (a) lim f x 2 42 3 4 1 2127. lim sin x sinx 2x 4 12321 6 3(b) lim g x x 21(c) lim g f x g 21 3x 4 x2 1 cos 33228. lim tan x tan 029. lim cos31. lim sec 2x sec 0 132. lim cos 3x cos 3 1x x 034.lim cos x cosx 5 35 1 32x 2x 35. lim tanx 3 4x tan 34 130. lim sinx 133. x sin 122lim sin x sinx 5 636. lim secx 75 1 62 6x sec 76 23 3
Section 1.337. (a) lim 5g x 5 lim g x 5 3 15x cx c(b) lim f x g x lim f x lim g x 2 3 5x cx cx cEvaluating Limits Analytically38. (a) lim 4f x 4 lim f x 4x cx c(c) lim f x g x lim f x lim g x 2 3 6x cx c(d) limx cx clim f x f x 2x c g x lim g x 3 32 6(b) lim f x g x lim f x lim g x x cx cx c(c) lim f x g x lim f x lim g x x cx c59x c3 1 22 2 32 12 43lim f x 3 2f x x c 3 x c g x lim g x 1 2x c(d) limx c39. (a) lim f x 3 lim f x 3 4 3 64x cx c333 lim f x 27 3f x 40. (a) lim x cx clim f x 27 3f x x c 18lim 1818 2(b) lim f x lim f x 4 2(b) lim(c) lim 3 f x 3 lim f x 3 4 12(c) lim f x lim f x 2 27 2 729x cx cx cx cx cx c(d) lim f x x c3 22x c lim f x 3 2x c 4 3 2 8x c(d) lim f x 2 3x c41. f x 2x 1 and g x x 0. 2x2 xagree except atx lim f x 2 3 27 2 3 9x c42. f x x 3 and h x x2 3xagree except at x 0.x(a) lim g x lim f x 1(a) lim h x lim f x 5(b) lim g x lim f x 3(b) lim h x lim f x 3x 0x 2x 0x 1x 0x 143. f x x x 1 and g x x3 xagree except at x 1.x 144. g x x 2x 01xand f x 2agree except at x 0.x 1x x(a) lim g x lim f x 2(a) lim f x does not exist.(b) lim g x lim f x 0(b) lim f x 1x 1x 1x 145. f x x 1x2 1and g x x 1 agree except at x 1.x 1lim f x lim g x 2x 1x 1x 02x2 x 3and g x 2x 3 agree except atx 1x 1.46. f x x 1lim f x lim g x 5x 13x 14 34 84 4 8
60Chapter 147. f x x 2.Limits and Their Propertiesx3 8and g x x2 2x 4 agree except atx 2x3 1and g x x2 x 1 agree except atx 1x 1.48. f x lim f x lim g x 12x 2lim f x lim g x 3x 2x 1x 1127 9 494 10x 5x 5 limx2 25 x 5 x 5 x 5 49. limx 5 limx 551. limx 3 x 5 5x lim 2 x 2x limx 0x 4 x 5 3x 4 limx 4 limx 456. limx 3 x 1 2x 3 limx 3 x 5 5 limx 4 x 5 51 lim 2 x 2 2 x 2 x2 x 2x 4 x 5 5x 01 2 5 510 2 x 2 2 x 2 x x 5 3 x 1 3 1 x 2 6 2 x 5 5x x 5 5x 0 lim55. limx 4x 11 x 24x2 5x 4 x 4 x 1 limx2 2x 8 x 4 x 4 x 2 52. lim x 5 5x 0x 0x 0x 2x 2 5 5 x 3 6 6 lim54. lim limx2 x 6 x 3 x 2 limx 3 x 3 x 3 x2 9x 3x 0x 211 x 5 10 lim53. lim2 x x 2 limx2 4 x 2 x 2 x 2 50. lim limx 01 2 x 2 21 42 2 x 5 3 x 5 3 x 5 9111 lim x 4 36 x 5 3 9 x 4 x 5 3 x 1 2x 3 x 1 2 x 1 2 limx 311x 3 lim x 3 x 1 2 x 3 x 1 2411 3 x 33 3 x x 1157. lim lim lim lim x 0x 0 3 x 3 x x 0 3 x 3 x x 0 3 x 3x9114 x 4 x 4 44 x 4 lim58. limx 0x 0xx limx 0 11 4 x 4 1659. lim x 02 x x 2x2x 2 x 2x lim x 0 x x lim 2 2 x 0
Section 1.360. lim x 0Evaluating Limits Analytically x x 2 x2x2 2x x x 2 x2 x 2x x lim lim lim 2x x 2x x 0 x 0 x 0 x x x x x 2 2 x x 1 x2 2x 1 x2 2x x x 2 2x 2 x 1 x2 2x 1 lim x 0 x 0 x x61. lim lim 2x x 2 2x 2 x 062. lim x 0 x x 3 x3x3 3x2 x 3x x 2 x 3 x3 lim x 0 x x lim x 063. lim x 2 2x 0xf x x 0.1 0.354 0.010.358Analytically, lim x 3x2 3x x x 2 lim 3x2 3x x x 2 3x2 x 0 x200.0010.010.10.354?0.3540.3530.3490.354 x 2 2xx 0 lim x 2 2xx 0 limx 064. f x 3 0.001 3 2 x 2 2 x 2 2 2x 2 211 0.354. lim x 0 4 x 2 2x x 2 2 2 24 xx 1610x15.915.9915.9991616.00116.0116.1f x 0.1252 0.125 0.125? 0.125 0.125 0.124820 1It appears that the limit is 0.125.4 x 4 x limx 16 x 16x 16 x 4 x 4 Analytically, lim limx 16 1 x 41 .811 12 x 265. lim x 0x43 5x 0.1 0.01 0.00100.001f x 0.263 0.251 0.250? 0.25011 2 x 22 2 x Analytically, lim limx 0x 0x2 2 x 10.010.1 0.249 0.238 x11 2 11lim . x x 0 lim2 2 x x x 0 2 2 x 461
62Chapter 1Limits and Their Propertiesx5 32 80x 266. limx 2x1.9f x 1001.9972.3979.20Analytically, limx 180.802.1 43 2588.41x5 32 x 2 x4 2x3 4x2 8x 16 limx 2x 2x 2 lim x4 2x3 4x2 8x 16 80.x 2(Hint: Use long division to factor x5 32.)67. limx 069. limx 0sin x limx 05x sinx x 15 sin x 1 cos x 1 limx 0 22x2 1 15 51sin xx 68. lim3 1 cos x 1 cos x lim 3x 0xx70. limcos tan sin lim 1 0 x 01 cos xx 0 3 0 01 1 0 02tan2 xsin2 xsin x lim lim2x 0x 0 x cos xx 0xxsin2 xsin x limsin x 1 sin 0 0x 0x 0xx71. lim72. limsin x cos2 x 1 0 0 1 cos h 21 cos h lim 1 cos h h 0h 0hh74. lim sec 1 73. lim 0 0 075. limx 276. limx 4cos x lim sin x 1x 2cot x1 tan xcos x sin x limsin x cos x x 4 sin x cos x cos2 x sin x cos x limx 4 cos x sin x cos x 1 limx 4 cos x77. limt 0 lim sec x x 4 278. limx 0 sin 2xsin 2x lim 2sin 3x x 02x 13 sin3x3x 2 1 13 1 32 sin 3tsin 3t limt 02t3t 32 1 32 23
Section 1.379. f t sin 3tt4 0.1tEvaluating Limits Analyticallyf t 0.012.96 0.0012.99963 00.0010.010.1?32.99962.9680. From the graph, limx 02 1The limit appears to equal 3. sin 3tsin 3t lim 3 3 1 3.t 0t 0t3tAnalytically, lim 2 cos x 1 0.252x21 1 0.1 0.010.010.11 0.2298 0.2498 0.25 0.25 0.2498 0.2298xf x Analytically,cos x 12x2cos x 1cos2 x 1 cos x 1 2x2 cos x 1 sin2 x cos x 1 2x2sin2 xx2 1 2 cos x 1 sinx x 2 cos 1x 1 1 14 4 0.252lim 1cos x 1 0.252x2limx 0x 0 1281. f x sin x2x1 2 x 0.1 0.01 0.00100.0010.010.1f x 0.099998 0.01 0.001?0.0010.010.099998Analytically, limx 082. f x 2 1 sin x2sin x2 lim x 0 1 0.x 0xx2sin x3 x2 3 0.1xf x 0.215 0.010.0464 0.00100.0010.010.10.01?0.010.04640.215 sin xsin x3 2Analytically, lim 3 lim x 0 1 0.x 0 xx 0x83. lim x 03 2The limit appear to equal 0.f x x f x 2 x x 3 2x 3 2x 2 x 3 2x 32 x lim lim 2 lim x 0 x 0 x 0 x x x x63
64Chapter 184. lim x 0Limits and Their Properties x x x x x xf x x f x lim lim x 0 x 0 x x x lim x 0 x x x x x xx x x11 lim x 0 x x x x x x x 2 x44 4f x x f x x xx4x 4 x x 4 lim 2 lim85. lim lim x 0 x 0 x 0 x x x x x 0 x x x x xxf x x f x x x 2 4 x x x2 4x x2 2x x x2 4x 4 x x2 4x lim lim x 0 x 0 x 0 x x x86. lim lim x 0 x 2x x 4 lim 2x x 4 2x 4 x 0 x87. lim 4 x2 lim f x lim 4 x2 x 0x 0 88. lim b x ax ax 0 Therefore, lim f x b.Therefore, lim f x 4.x ax 0 90. f x x sin x89. f x x cos x91. f x x sin x6643 2 2 2 2 2 4 6 2 x 0 93. f x x sin92. f x x cos x6 lim x sin x 0lim x sin x 0lim x cos x 0x 0x 01x94. h x x cos0.52 0.5 0.50.5 0.5 lim x cos x 0 lim x sinx 01x0.50.5 6x 0x ax ax 0 2 x ab lim f x b4 lim f x 4 3 2 lim f x lim b x a 0.5 1 0x lim x cosx 0 1 0xx2 1and g x x 1 agree at all pointsx 1except x 1.95. We say that two functions f and g agree at all but onepoint (on an open interval) if f x g x for all x in theinterval except for x c, where c is in the interval.96. f x 97. An indeterminant form is obtained when evaluating a limitusing direct substitution produces a meaningless fractionalexpression such as 0 0. That is,98. If a function f is squeezed between two functions h and g,h x f x g x , and h and g have the same limit L asx c, then lim f x exists and equals L.f x limx c g x for which lim f x lim g x 0x cx cx c
Section 1.399. f x x, g x sin x, h x sin xxEvaluating Limits Analytically100. f x x, g x sin2 x, h x 365sin2 xx2fghg 55 33hf 3 2When you are “close to” 0 the magnitude of f isapproximately equal to the magnitude of g.Thus, g f 1 when x is “close to” 0.When you are “close to” 0 the magnitude of g is“smaller” than the magnitude of f and the magnitudeof g is approaching zero “faster” than the magnitude of f.Thus, g f 0 when x is “close to” 0. 101. s t 16t2 1000limt 5s 5 s t 600 16t2 1000 16 t 5 t 5 lim lim lim 16 t 5 160 ft sec.t 5t 5t 55 t5 t t 5 Speed 160 ft sec102. s t 16t2 1000 0 when t slimt 5 10 2 5 210 s t 5 10 seconds 1000162 5 10 t2 0 16t2 1000 t 5 10 25 10 t2lim 16 t2 limt 5 10 2limt 5 10 212525 10 t2 16 t t 5 210 t 5 10 216 t limt 5 10 25 102 5 10 80 10 ft sec 253 ft sec2103. s t 4.9t2 150s 3 s t 4.9 32 150 4.9t2 150 4.9 9 t2 lim limt 3t 3t 33 t3 t3 tlim limt 3 4.9 3 t 3 t lim 4.9 3 t 29.4 m sect 33 t104. 4.9t2 150 0 when t 1500 5.53 seconds. 1504.949The velocity at time t a iss a s t 4.9a2 150 4.9t2 150 4.9 a t a t lim limt at at aa ta ta tlim lim 4.9 a t 2a 4.9 9.8a m sec.t aHence, if a 1500 49, the velocity is 9.8 1500 49 54.2 m sec.
66Chapter 1Limits and Their Properties105. Let f x 1 x and g x 1 x. lim f x and lim g x do not exist.x 0x 0 11lim f x g x lim x 0x 0 xx lim 0 0x 0106. Suppose, on the contrary, that lim g x exists. Then, sincex clim f x exists, so would lim f x g x , which is ax cx ccontradiction. Hence, lim g x does not exist.108. Given f x x n, n is a positive integer, thenx cx c f x L b whenever 0 x c wherever 0 x c , we have b f x L or bf x bL x c c lim x lim xn 2 c c lim xxn 3 x c x c lim x lim x n 1 c lim x x n 2 x c 109. If b 0, then the property is true because both sidesare equal to 0. If b 0, let 0 be given. Sincelim f x L, there exists 0 such thatlim x n lim x x n 1 x c x cx c107. Given f x b, show that for every 0 there exists a 0 such that f x b whenever x c .Since f x b b b 0 for any 0,then any value of 0 will work.x c . . . c n. Hence,which implies that lim bf x bL.x c110. Given lim f x 0: x c lim M f x lim f x g x lim M f x For every 0, there exists 0 such thatf x 0 whenever 0 x c .Now f x 0 f x f x 0 forx c . Therefore, lim f x 0. M f x f x g x M f x 111.x cx cx cx c M 0 lim f x g x M 0 x c0 lim f x g x 0x cTherefore, lim f x g x 0.x c f x f x f x f x lim f x lim f x x cx c(b) Given lim f x L:112. (a) If lim f x 0, then lim f x 0.x climx cx cx cFor every 0, there exists 0 such thatf x L whenever 0 x c .Since f x L f x L forx c , then lim f x L . 0 lim f x 0x c x cTherefore, lim f x 0.x c113. False. As x approaches 0 from thexleft, 1.x 114. False. limx sin x0 0x 115. True2 33 2116. False. Letf x 117. False. The limit does not exist.x x 1,3 x 1c 1.Then lim f x 1 but f 1 x 11. 36 2118. False. Letf x 12x2 and g x x2.4Then f x g x for all x 0.But lim f x lim g x 0.x 0x 0
Section 1.3120. lim119. Letx 0f x 4, 4, 1 cos x1 cos x limx 0xxif x 0if x 0671 cos x 1 cos x1 cos2 xsin2 x limx 0 x 1 cos x x 0 x 1 cos x limlim f x lim 4 4.x 0Evaluating Limits Analytically limx 0x 0lim f x does not exist since for x 0, f x 4 and forx 0, f x 4.x 0sin xx limx 0sin x 1 cos xsin xxlimx 0sin x1 cos x 1 0 0121. f x 0, if x is rational1, if x is irrationalg x 0, if x is rationalx, if x is irrational122. f x sec x 1x20, 2 n .(a) The domain of f is all x(b)2lim f x does not exist.x 0 3 2No matter how “close to” 0 x is, there are still an infinitenumber of rational and irrational numbers so that lim f x x 0does not exist. 2lim g x 0The domain is not obvious. The hole at x 0 is notapparent.x 0When x is “close to” 0, both parts of the function are“close to” 0.3 2(c) lim f x x 0(d)12sec x 1 sec x 1 x2x2 Hence, limx 0sec x 1 x 01 cos x1 cos x limx 0x2x21 cos x 1 cos x lim1 cos2 xx2 1 cos x limsin2 xx2x 0x 0 1 1 1 cos x 12 21124. The calculator was set in degree mode, instead of radian mode.(b) Thus, 1sin2 x1tan2 x 2 sec x 1 cos x x2 sec x 1x2sec x 11 sin2 x1 lim2x 0xcos2 x x2 sec x 1 1 1 123. (a) limsec2 x 1 sec x 1 x2 sec x 1 12 21.
48 Chapter 1 Limits and Their Properties 4. 1 lim Actual limit is 4. x 3 1 x 2 x 3 0.25 x 3.1 3.01 3.001 2.999 2.99 2.9 f x 0.2485 0.2498 0.2500 0.2500 0.2502 0.2516 5.
